Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
Подождите немного. Документ загружается.
SECTION
1.5
Equivalence
From the borrower's perspective, the rate
of
inflation
is
simply another interest
rate
tacked on
to
the real interest rate. And, from the vantage point
ofthe
saver or
investor in a fixed-interest account, inflation
reduces the real rate
of
return on the
investment. Inflation means that cost and revenue cash flow estimates increase
over time. This increase is due to the changing value
of
money that
is
forced upon
a country's currency
by
inflation, thus making a unit
of
currency (one dollar) worth
less relative to its value at a previous time. We see the effect
of
inflation in that
money purchases less now than it did at a previous time. Inflation contributes to
• A reduction in purchasing power
of
the currency.
• An increase in the CPI (consumer price index).
• An increase in the cost
of
equipment and its maintenance.
• An increase
in
the cost
of
salaried professionals and hourly employees.
A reduction
in
the real rate
of
return on personal savings and certain corporate
investments.
In other words, inflation can materially contribute to changes in corporate and
personal economic analysis.
Commonly, engineering economy studies assume that inflation affects all es-
timated values equally. Accordingly, an interest rate or rate
of
return, such as 8%
per year,
is
appljed throughout the analysis without accounting for an additional
inflation rate. However, if inflation were explicitly taken into account, and it was
reducing the value
of
money at, say, an average
of
4% per year, it would be nec-
essary to perform the economic analysis using an inflated interest rate
of
12.32%
per year. (The relevant relations are derived
in
Chapter 14.) On the other hand,
if
the stated
ROR
on an investment is 8% with inflation included, the same inflation
rate
of
4% per year results in a real rate
of
return
of
only 3.85% per year!
1.5
EQUIVALENCE
Equivalent terms are used very often
in
the transfer from one scale to another.
Some common equivalencies or conversions are as follows:
Length:
100 centimeters = 1 meter 1000 meters = 1 kilometer
12 inches
= I foot 3 feet = 1 yard 39.370 inches = 1 meter
Pressure:
I atmosphere = 1 newton/meter
2
1 atmosphere = 10
3
pascal = 1 kilopascal
Many equivalent measures are a combination
of
two or more scales. For example,
110
kilometers per hour (kph)
is
equivalent to 68 miles per hour (mph) or
1.133 miles per minute, based on the equivalence that 1
mile = 1.6093 kilometers
and
I hour = 60 minutes.
We
can further conclude that driving at approxi-
mately 68 mph for 2 hours
is
equivalent to traveling a total
of
about 220 kilome-
ters, or 136 miles. Three
scales-time
in
hours, length in miles, and length in
kilometers-are
combined to develop equivalent statements. An additional use
of
these equivalencies is to estimate driving time in hours between two cities using
two maps, one indicating distance in miles, a second showing kilometers. Note that
throughout these statements the fundamental relation 1
mile = 1.6093 kilometers
is used.
If
this relation changes, then the other equivalencies would be in error.
·
15
16
Figure
1-3
Equivalence
of
three
amounts at a 6% per year
interest rate.
CHAPTER 1
Foundations
of
Engineering Economy
6% per year interest rate
$94.34 I $5.66 I $6.
00
0
$
100
.
00
II $6.00 0
I year ago Now
$106.00 0
I year
from now
When considered together, the time value
of
money and the interest rate help
develop the concept
of
economic equivalence, which means that different sums
of
money at different times would be equal in economic value.
For
example, if
the interest rate
is
6% per year, $100 today (present time)
is
equivalent to $106
one year from today.
Amount accrued
= 100 + 100(0.06) = 100(1 + 0.06) = $106
So,
if
someone offered you a gift
of
$100 today or $106 one year from today, it
would make no difference which offer you accepted from
an
economic perspective.
In either case you have
$106 one year from today. However, the two sums
of
money
are equivalent to each other
only when the interest rate
is
6% per year. At a higher
or lower interest rate,
$100 today is not equivalent to $106 one year from today.
In addition to future equivalence, we can apply the same logic to determine
equivalence for previous years. A total
of
$100 now is equivalent to
$100/1.06
=
$94.34 one year ago at an interest rate
of
6% per year. From these illustrations, we
can state the following: $94.34 last year,
$100 now, and $106 one year from now
are equivalent at an interest rate
of
6% per year. The fact that these sums are equiv-
alent can be verified by computing the two interest rates for I-year interest periods.
$6
$100 X 100% = 6% per year
and
$5.66
$94.34
X 100% = 6% per year
Figure 1
-3
indicates the amount
of
interest each year necessary to make these
three different amounts equivalent at 6% per year.
EXAMPLE
1.6
",;;;t;
,
AC-
De
lco makes auto batteries avaiJable to General Motors dealers through privately
owned distributorship
s.
In general, batteries are stored throughout the year, and a 5%
cost increase is added each year to cover the inventory carrying char
ge
for the distribu-
torship owner. Assume you own the City Center Delco facility. Make the calculations
SECTION
1.6
Simple and Compound Interest
necessary to show which of the following statements are true and which are false about
battery costs.
(a) The amount
of
$98 now
is
equivalent
to
a cost
of
$105.60 one year from
now.
(b) A truck battery cost
of
$200 one year ago
is
equivalent to $205
now.
(c) A $38 cost now
is
equivalent
to
$39.90 one year from
now.
(d)
A $3000 cost now
is
equivalent to $2887.
14
one year ago.
(e)
The carrying charge accumulated in 1 year on
an
investment
of
$2000 worth
of
batteries
is
$100.
Solution
(a) Total amount accrued = 98(1.05) = $102.90
"*
$105.60; therefore, it
is
fals
e.
Another way
to
solve this
is
as follows: Required original cost
is
105.60/1.05 =
$100.57
"*
$98.
(b) Required old cost is 205.00/1.05 = $195.24
"*
$200; therefore, it
is
false.
(c) The cost 1 year from now is $38(1.05) = $39.90; true.
(d)
Cost now is 2887.14(1.05) = $3031.50
"*
$3000; false.
(e)
The charge
is
5% per year interest, or 2000(0.05) = $100; true.
1.6
SIMPLE
AND
COMPOUND
INTEREST
The terms interest, interest period, and interest rate (introduced in Section 1.4)
are useful in calculating equivalent sums
of
money for one interest period in the
past and one peliod in the future. However, for more than one interest period, the
terms
simple interest and compound interest become important.
Simple interest
is
calculated using the principal only, ignoring any interest ac-
crued
in
preceding interest periods. The total simple interest over several periods
is
computed as
Interest
=
(principal)(number
of
periods)(interest rate) [1.5]
where the interest rate is expressed
in
decimal form.
EXAMPLE
1.7
~.
'
Pacific Telephone Credit Union loaned money
to
an
engineering staff member for a radio-
controlled model airplane. The loan
is
for
$1000 for 3 years at 5% per year simple inter-
est. How much money will the engineer repay at the end of 3 years? Tabulate the results.
Solution
The interest for each
of
the 3 years is
Interest per year
= 1000(0.05) = $50
Total interest for 3 years from Equation [1.5]
is
Total interest = 1000(3)(0.05) = $150
The amount due after 3 years
is
Total due = $1000 + 150 =
$1
150
17
18
CHAPTER 1 Foundations of Eng
in
eering Economy
The
$50 interest accrued
in
the first year and the $50 accrued
in
the second year do
not earn interest. The interest due each year is calculated only
on
the $1000 principal.
The details of this loan repayment are tabulated
in
Table
I-I
from the perspective
of
the borrower. The year zero represents the present, that
is
, when the money
is
bon·owed.
No payment
is
made until the end
of
year 3. The amount owed each year increases uni-
formly by
$50, since simple if)terest is figured on only the loan principal.
TABLE
1- 1
Si
mple Interest Computations
(1
)
(2)
(3)
(4)
(5)
End
of
Amount
Amount
Amount
Year
Borrowed
Interest
Owed
Paid
0 $1000
I $50 $1050 $ 0
2 50 1100 0
3 50
1150 1150
For compound interest, the interest accrued for each interest period is calculated
on the
principal plus the total amount
of
interest accumulated in all previous
periods.
Thus, compound interest means interest on top
of
interest. Compound
interest reflects the effect
of
the time value
of
money on the interest also. Now
the interest for one period
is
calculated
as
Interest
= (principal + all
accrued
interest)(interest rate) [1.6]
EXAMPLE
1.8
'OitllJ;
If
an
engineer borrows $1000 from the company credit union at 5% per year compound
interest, compute the total amount due after 3 years. Graph and compare the results
of
this and the previous example.
Solution
The interest and total amount due each year are computed separately using Equa-
tion [1.6].
Year 1 interest:
Total amount due after year
1:
Year 2 intere
st:
Total amount due after year 2 :
Year 3 interest:
Total amount due after year
3:
$1000(0.05) = $50.00
$1000
+ 50.00 = $1050.00
$1050(0.05)
= $52.50
$1050 + 52.50 = $1102.50
$1102.50(0.05)
= $55.13
$1102.50
+ 55.13 = $1157.63
SECTION 1.6 Simple and Compound Interest
TABLE
1-2
Compound Interest Computations, Example
1.8
(1)
(2)
(3)
(4)
(5)
End
of
Amount Amount Amount
Year
Borrowed
Interest
Owed
Paid
0 $1000
1
$50.00 $1050.00
$ 0
2 52.50 1102.50
0
3 55.13 1157.63 1157.63
The details are shown
in
Table
1-2
. The repayment plan is the same as that for the sim-
ple interest
example-no
payment until the principal plus accrued interest is due at the
end
of
year
3.
Figure 1-4 shows the amount owed at the end
of
each year for 3 years. The differ-
ence due
to
the time value
of
money
is
recognized for the compound interest case.
An
extra $1157.63 - $1150 = $7.63
of
interest is paid compared with simple interest over
the 3-year period.
0
V)
N
0
0
-
S
-
""
0 0
fA-
,,..,
V)
r-
0 0
-
-
""
""
-,--
S C
S
C
0
V)
N
0 0
0
V)
V) V)
V)
""
""
""
""
C
2
End
of
year
Figure
1-4
0
,--
V)
-
-
""
r-
S
C
~
vi
on
""
0
V)
""
S C
3
S - Simple
Interest
C
-
Compound
Int
erest
Comparison
of
simp
le and compound interest calculations, Examples 1.7 and 1.8.
1.9
20
CHAPTER 1
Foundations
of
Engineering Economy
Comment
The difference between simple and compound interest grows each year.
If
the compu-
tations are continued for more years, for example, 10 years, the difference
is
$128.90;
after
20 years compound interest is $653.30 more than simple interest.
If
$7.63 does not seem like a significant difference
in
only 3 years, remember that
the beginning amount here is
$1000.
If
we make these same calculations for
an
initial
amount
of
$100,000 or
$1
million, multiply the difference
by
100 or 1000, and
we
are
talking real money. This indicates that the power
of
compounding is vitally important
in
all
economics-based analyses.
Another and shorter way to calculate the total amount due after 3 years in
Example
1.8 is to
combine
calculations rather than perform them on a year-by-
year basis.
The
total due each year is as follows:
Year
1:
Year 2:
Year
3:
$1000(1.05)1 = $1050.00
$1000(1.05)2 = $1102.50
$1000(1.05)3 = $1157.63
The
year 3 total is calculated directly; it does not require the year 2 total. In gen-
eral formula form
Total due after a number
of
years = principal
(l
+ interest
rate)"ull1b
e
rof
y
e
ar
s
This
fundamental relation is used many times in upcoming chapters.
We
combine
the concepts
of
interest rate, simple interest,
compound
interest,
and equivalence to demonstrate that different loan repayment plans may be
equivalent, but differ substantially in monetary amounts from
one
year to an-
other. This also
shows
that there are many ways to take into account the
time
value
of
money.
The
following
example
illustrates equivalence for five different
loan repayment plans.
EXAMPLE
1.9
:.\
(a) Demonstrate the concept
of
equivalence using the different loan repayment
plans described below. Each plan repays a
$5000 loan
in
5 years at 8% interest
per year.
•
Plan
1: Simple interest,
pay
all
at
end. No interest or principal is paid until
the end
of
year
5.
Interest accumulates each year on the principal only.
• Plan 2:
Compound
interest,
pay
all
at
end. No interest or principal is paid
until the end
of
year 5. Interest accumulates each year on the total of princi-
pal and all accrued interest.
•
Plan
3: Simple interest paid annually, principal repaid
at
end. The accrued
interest
is
paid each year, and the entire principal
is
repaid at the end
of
year
5.
•
Plan
4:
Compound
interest
and
portion
of
principal
repaid
annually.
The accrued interest and one-fifth
of
the principal (or $1000) is repaid each
SECTION
1.6
Simple and Compound Interest
year. The outstand
in
g loan balance decreases each year, so the interest for
each year decreases.
•
Plan
5:
Equal
payments
of
compound interest
and
principal
made
annu-
ally. Equal payments are made each year with a portion going toward prin-
cipal repayment and the remainder covering the accrued interest. Since the
loan balance decreases at a rate slower than that
in
plan 4 due
to
the equal
end-of-year payments, the interest decreases, but at a slower rate.
(b) Make a statement about the equivalence
of
each plan at 8% simple or compound
interest,
as
appropriate.
Solution
(a) Table 1
-3
presents the interest, payment amount, total owed at the end
of
each
year, a
nd
total amount paid over the 5-year period (column 4 totals
).
TABLE
1-3
Different Repayment Schedules Over 5 Years for $5000 at
8%
Per Year Interest
(1
)
(2) (3)
(4)
(5)
End
of
Interest
Owed
Total
Owed
at
End-of-Year Total
Owed
Year for Year End
of
Year
Payment
after
Payment
Plan
I:
Simple Interest, Pay All at End
0
$5000.00
I $400.00 $5400.00
5400.00
2 400.00 5800.00 5800.00
3
400.00 6200.00
6200.00
4 400.00 6600.00 6600.00
5
400.00 7000.00 $7000.00
Totals $7000.00
Plan 2: Compound Inte
re
st,
Pa
y All at End
0
$5000.00
I
$400.00 $5400.00
5400.00
2 432.00 5832.00
5832.00
3
466.56 6298.56
6298.56
4
503.88 6802.44
6802.44
5
544
.2
0 7346.64 $7346.64
Totals $7346.64
Plan 3: Simple interest Paid Annually; Principal Repaid
at
End
0
$5000.00
$400.00 $5400.00 $400.00 5000.00
2 400.00 5400.00 400.00 5000.00
3 400.00 5400.00 400.00
5000.00
4 400.00 5400.00
400.00 5000.00
5 400.00 5400.00
5400.00
Totals
$7000.00
22
CHAPTER I Foundations
of
Engineering Economy
TABLE
1-3
(Continued)
(1) (2) (3) (4) (5)
End
of
Interest
Owed
Total
Owed
at
End-of-Year Total
Owed
Year for Year
End
of
Year Payment after Payment
Plan 4: Compound Interest and Portion
of
Principal Repaid Annually
o
~~OO
$400.00 $5400.00 $1400.00 4000.00
2 320.00 4320.00 1320.00 3000.00
3 240.00 3240.00 1240.00 2000.00
4 160.00 2160.00 1160.00 1000.00
5 80.00 1080.00 1080.00
Totals $6200.00
Plan
5: Equal Annual Payments
of
Compound Interest and Principal
o $5000.00
$400.00
$5400.00
$1252.28
2
331.82
4479.54 1252.28
3
258.18 3485.43 1252.28
4
178.65
2411.80 1252.28
5
92.76 1252.28 1252.28
Totals $6261.41
The amounts
of
interest (column 2) are determined
as
folJows:
Plan 1
Simple interest = (original principal)(0.08)
Plan 2 Compound interest
= (total owed previous year)(0.08)
Plan 3
Simple interest = (original principal)(0.08)
Plan 4 Compound interest
= (total owed previous year)(0.08)
Plan 5 Compound interest
= (total owed previous year)(0.08)
4147.72
3227.25
2233.15
1159.52
Note that the amounts
of
the annual payments are different for each repayment sched-
ule and that the total amounts repaid for most plans are different, even though each
repayment plan requires exactly 5 years. The difference
in
the total amounts repaid can
be
explained
(I)
by
the time value
of
money, (2) by simple or compound interest, and
(3)
by
the partial repayment
of
principal prior to year
5.
(b) Table 1
-3
shows that $5000 at time 0
is
equi valent to each
of
the
fol
lowi
ng:
Plan 1 $7000 at the end
of
year 5 at
8%
simple interest.
Plan 2 $7346.64 at the end
of
year 5 at 8% compound interest.
Plan 3
$400 per year for 4 years and $5400 at the end
of
year 5 at 8% simple interest.
Plan 4 Decreasing payments
of
interest and partial princi
pal
in
years I
($
I 400)
through 5 ($1080) at 8% compound interest.
Plan 5 $1252.28 per year for 5 years at 8% compound interest.
An
engineering economy study uses plan 5; interest
is
compounded, and a constant
amount
is
paid each period. This amount covers accrued interest and a partial amount
of
principal repayment.
SECTION 1.7 Terminology and Symbols
1.7 TERMINOLOGY
AND
SYMBOLS
The equations and procedures
of
engineering economy utilize the following
terms and symbols. Sample units are indicated.
P = value
or
amount
of
money at a time designated as the present
or
time
O.
Also P is referred to as present worth (PW), present value
(PV), net present value (NPV), discounted cash flow (DCF), and
capitalized cost (CC); dollars
F = value or amount
of
money at some future time. Also F is called
future worth (FW) and future value (FV); dollars
A = series
of
consecutive, equal, end-of-period amounts
of
money.
Also
A is called the annual worth (AW) and equivalent uniform
annual worth
(EUA W); dollars per year, dollars per month
n = number
of
interest periods; years, months, days
i = interest rate or rate
of
return per time period; percent per year,
percent per month; percent per day
t = time, stated in periods; years, months, days
The symbols
P and F represent one-time occurrences: A occurs with the same
value once each interest period for a specified number
of
periods.
It
should be
clear that a present value
P represents a single sum
of
money at some time
prior to a future value
F or prior to the first occurrence
of
an equivalent series
amountA.
It
is
important to note that the symbol A always represents a uniform
amount (i.e., the same amount each period) that extends through
consecutive
interest periods. Both conditions must exist before the series can be repre-
sented by
A.
The interest rate i
is
assumed to be a compound rate, unless specifically stated
as simple interest. The rate
i is expressed in percent per interest period, for
example, 12% per year. Unless stated otherwise, assume that the rate applies
throughout the entire
n years or interest periods. The decimal equivalent for i is
always used
in
engineering economy computations.
All engineering economy problems involve the element
of
time n and interest
rate
i.
In general, every problem will involve at least four
of
the symbols P, F,
A,
n, and i, with at least three
of
them estimated
or
known.
EXAMPLE
1.10
e
A new college graduate has a
job
with Boeing Aerospace. She plans to
bOlTOW
$10,000
now
to
help
in
buying a car. She has arranged to repay the entire principal plus 8%
per year interest after 5 years. Identify the engineering economy symbols involved and
their values for the total owed after 5 years.
23
24
CHAPTER I Foundations
of
Engineering Economy
Solution
In this case, P and F are involved, since all amounts are single payments,
as
well
as
n
and i. Time is expressed in years.
P = $10,000 i = 8% per year n = 5 years
F=?
The future amount F
is
unknown.
Assume you borrow
$2000 now at 7% per year for
10
years and must repay the loan
in
equal yearly payments. Determine the symbols involved and their value
s.
Solution
Time
is
in
years.
P = $2000
A = ? per year for 5 years
i = 7% per year
n =
10
years
In Examples 1.10 and
1.11
, the P value
is
a receipt to the borrower, and
For
A
is
a disbursement from the borrower. It is equally correct to use these symbols
in
the reverse roles.
EXAMPLE
1.12
.'k
On
July
I,
2002, your new employer Ford Motor Company deposits $5000 into your
money market account,
as
part
of
your employment bonus. The account pays interest at
5% per year.
You
expect to withdraw
an
equal annual amount for the following
10
years. identify the symbols and their values.
Solution
Time is
in
years.
P = $5000
A = ? per year
i = 5% per year
n =
10
years